Circuits and Systems, 2011, 2, 297-310
doi:10.4236/cs.2011.24042 Published Online October 2011 (http://www.scirp.org/journal/cs)
Distortionless Lossy Transmission Lines Terminated by in
Series Connected RCL-Loads
Vasil G. Angelov1, Marin Hristov2
1
Department of Mathematics, University of Mining and Geology, Sofia, Bulgaria
Department of Microelectronics, Technical University of Sofia, Sofia, Bulgaria
E-mail: angelov@mgu.bg, mhristov@ecad.tu-sofia.bg
Received July 29, 2011; revised August 26, 2011; accepted September 4, 2011
2
Abstract
The paper deals with a lossy transmission line terminated at both ends by non-linear RCL elements. The
mixed problem for the hyperbolic system, describing the transmission line, to an initial value problem for a
neutral equation is reduced. Sufficient conditions for the existence and uniqueness of periodic regimes are
formulated. The proof is based on the finding out of suitable operator whose fixed point is a periodic solution
of the neutral equation. The method has a good rate of convergence of the successive approximations even
for high frequencies.
Keywords: Lossy Transmission Line, RCL-Nonlinear Loads, Fixed Point, Periodic Solution
1. Introduction
The principal importance of transmission lines investigations has been discussed in many papers (cf. for
instance [1-8]). In a previous paper [9] we have investigated lossless transmission lines terminated by in series
connected RCL-loads. In [10] we have considered a lossy
transmission line terminated by a resistive load with
exponential V-I characteristic. In [11] we have considered periodic regimes for lossy transmission lines terminated by parallel connected RCL-loads. Here we investigate lossy transmission lines terminated at both ends
by in series connected RCL-loads but in contrast of [11]
the capacitive element has a nonlinear V-C characteristic.
Unlike of the usually accepted approach (cf. for instance
[12,13]) we consider first order hyperbolic system instead of the Telegrapher’s equation derived from it. First
we reduce the mixed problem for the hyperbolic system
to an initial value problem for neutral system of equations on the boundary [14]. Extending ideas from [1517] we introduce operators whose fixed points are periodic solutions of the neutral system. Our treatment is
based on the fixed point method (cf. [18]). All derivation
are performed under assumption
R / L  G / C ( R  0, G  0 ).
The last condition is known as Heaviside one and it
implies that the waves propagate without distortion.
We would like to mention the advantages of our
Copyright © 2011 SciRes.
method in comparison of the other used ones: lumped
element method, finite element method and finitedifference time-domain method (cf. for instance [19-21]).
If we use numerical methods we have to keep one and
same accuracy. But here we consider nonlinearities of
polynomial and transcendental type (for exponential ones
cf. [10]). For such “bad” nonlinearities (cf. [2]) there are
examples showing that if we want to keep the same
accuracy it should be reduced step thousands of times.
Here we obtain (even though approximate) an analytical
solution for voltage and current beginning with simple
initial approximations.
We proceed from the system:
u  x, t  i  x, t 

 Gu  x, t   0,
t
x
i  x, t  u  x, t 
L

 Ri  x, t   0,
t
x
C
(1)
 x, t      x, t   R 2 :  x, t   0,    0,   ,
u x,0   u 0 x , i  x,0   i0  x  , x   0,  
(2)
where L, C, R and G are prescribed specific parameters
of the line and   0 is its length. Here the current
i  x, t  and voltage u  x, t  are unknown functions.
The initial conditions for the foregoing system (1) are
prescribed functions u 0 x , i0 x  . The boundary conditions can be derived from the loads and sources at the
ends of the line (cf. Figure 1).
CS
298
V. G. ANGELOV
ET AL.
Analogously for x   (cf. Figure 1) we obtain
t

u  , t   E1  t   R1  i  , t    C11   i  , t  d 
T

(5)

 di  , t 
dL1  i  , t  
.
 i   , t 
 L1  i  , t   
di

 dt
Figure 1. Lossy transmission line terminated by nonlinear
RCL-loads.
In view of the Kirchoff’s voltage law for the voltages
between the nodes a, b and d for x = 0 we obtain:
u  0, t   uad  uba  ubd  uba  E0  t  ,
(3)
where E0 (t ) is the source voltage. The voltage uba is:
uba  uR0  uC0  u 0  R0  i  0, t    uC0 
d 0
dt
.
To calculate the voltage of the condenser C0 we proceed from the relation (assuming
uC0 T   u T   0 ),
i
t
dq d  C0  u  .u 

  i  x,  d C0  u   u  C0  u 
dt
dt
T
or
t

uC0  u  0, t   C   i  0,  d  .
T

To calculate the voltage of the inductor L0 we proceed from
1
0
u 0
 dL  i 
 di
d 0

  L0  i   i    i 0
 L0  i  
dt
 di
 dt
.
Therefore the first boundary condition is:
t

uba  R0  i  0, t    C01   i  0,  d 
T


 di  0, t 
dL0  i  0, t  
 i  0, t 
 L0  i  0, t   
di

 dt
or
t

u  0, t   E0  t   R0  i  0, t    C01   i  0,  d 
T

(4)

 di  0, t 
dL0  i  0, t  
.
 i  0, t 
 L0  i  0, t   
di
 dt

Copyright © 2011 SciRes.
Here R1 . , C1 . and L1 . are characteristics (in
general, nonlinear functions) of RCL-loads at right end.
Now we are able to formulate the initial-boundary
(mixed) value problem for the hyperbolic transmission
line system of equations: to find a solution
system (1) for
 u  x, t  , i  x, t   of the hyperbolic
 x, t      x, t   R 2 : 0  x  , t  0 ,
satisfying the initial conditions

u  x,0   u0  x  , i  x,0   i0  x  for x   0, 
(6)
and the boundary conditions
t

u  0, t   E0  t   R0  i  0, t    C01   i  0,  d 
T

(7)

 di  0, t 
dL0  i  0, t  
 i  0, t 
 L0  i  0, t   
,
di

 dt
t

u  , t   E1  t   R1  i  , t    C11   i  , t  d 
T

(8)
dL1 (i  , t )

 di  , t 
 i   , t 
 L1  i  , t   
di

 dt
where u0  x  , i0  x  , Ek  t  , Rk . , Ck . and Lk (.)
(k  0, 1) are prescribed functions.
So the system (1) and conditions (6) - (8) form a mixed problem for a lossy transmission line equations.
2. Reducing the Mixed Problem for the
Transmission Line System to an
Initial Value Problem for a
Nonlinear Neutral System
First we present system (1) in matrix form:
U
U 

, Ux 
AU
1 t  A2U x  A3U  0  U t 

t
x 

(9)
C 0 
0 1
G 0 
where A1  
 , A2  1 0  , A3  0 R  ,
0
L






ut 
u x 
u 
U    , Ut    , U x    .
i 
it 
 ix 
Since A1  LC  0 , then multiplying Equation (9) by
A11 we obtain
CS
V. G. ANGELOV
U t  AU x  BU  0
(10)
 Vx 
Vt  1 LC 0
 
 
1 LC   I x 
 I t  0
0 1 C 
G C 0 
, B
where A  

 . In order to tran1 L 0 
0 R L 
0 1 C 
sform the matrix A  A11 A2  
 in a diagonal
1 L 0 
form we solve the characteristic equation:
 1 C
0
1 L 
whose roots are 1  1 LC and 2  1 LC . Denote the matrix formed by eigenvectors by
 C
L
H 

L 
  C
and its inverse one—by
1  1 C 1 C 
H 1  
.
2  1 L
1 L 
If we denote by
1 LC 0

Acan  
,
 1 LC 
 0
then Acan  HAH 1 .
Introduce new variables Z = HU (or U = H–1Z):
u  x, t   V  x , t  2 C  I  x , t  2 C
i  x, t   V  x , t  2 L  I  x , t  2 L .
(11)
(12)
Replacing U  H Z in Equation (10) we obtain

 H Z
  A H Z   B
1





1  G C    R L    G C    R L  
HBH 1  

2   G C    R L   G C    R L  
and then
Copyright © 2011 SciRes.

 1
Vt  1
It

LC  I
LC Vx   R L  V  0
x
  R L  I  0.
(14)
The new initial conditions we obtain from conditions
(6) and system (12):
V  x,0   C u  x, 0   L i  x, 0 
(15)
  C u0  x   L i0  x   I 0  x  , x   0,   .
(16)
Further on we set
 V  x, t   e
R L t
V  x, t  , J  x, t   e
 R L  t
R L t
W ( x, t ), I  x, t   e
I  x, t 
 R L  t
J  x, t  .
(17)

LC .
(18)
0
 x  , J  0, t   I  0, t   I 0  x  .
From system (12) and denotation (17) we obtain
u  x, t   W  x, t   J  x, t   e

But
or
W  0, t   V  0, t   V
Z t  H AH 1 Z x  H BH 1 Z  0 , i.e.
Z t  Acan Z x  HBH 1 Z  0 .
1
 Vx   R L 0  V  0 
  
  
LC   I x   0 R L   I   0 
The initial conditions (15), (16) remain the same ones:


0
LC

After multiplication from the left by H we obtain

Vt  1
 
 I t   0
Wt  vWx  0, J t  vJ x  0 v  1

H 1Z t  AH 1 Z x  BH 1 Z  0.

Then HBH–1 can be simplified and the last system becomes:
Then system (14) can be written in the form:
H 1 Z  0 .
t
x
Since H–1 is a constant matrix we have:

We consider distortionless lossy transmission lines
that means the following Heaviside condition is fulfilled:
R LG C.
W  x, t   e 
1
1
 G C    R L   V  0

.

 G C    R L    I  0
I  x, 0    C u  x , 0   L i  x , 0 
Then
I  x, t    C u  x , t   L i  x , t  ,
1  G C    R L 
 
2  G C    R L 
 C u0  x   L i0  x   V0  x  , x   0,   ,
V  x, t  
u  x, t  
Z 
.
, U  
 I  x, t  
i  x, t  
V  x, t   C u  x, t   L i  x , t 
299
ET AL.
(13)
  R L t
i  x, t   W  x, t   J  x, t   e  R L t
2 C ,
2 L 
(19)
and then
W  x, t   e
R Lt
J  x , t   e 
R Lt
C u  x, t   e
R Lt
C u  x, t   e
R Lt
L i  x, t  ,
L i  x, t  .
CS
300
V. G. ANGELOV
ET AL.
Replacing in system (19) x  0 we have
u  0, t   W  0, t   J  0, t   e
i  0, t   W  0, t   J  0, t   e
  R L t
  R L t
2 C ,
2 L 
and replacing x   —
u  , t   W  , t   J  , t   e
i  , t   W  , t   J  , t   e
2 C ,
  R L t
 2 L .
  R L t

(20)
(21)
Now replace expressions (20) into the first boundary
condition and obtain:
W  0, t   J  0, t   e

 E0  t   R0 W  0, t   J  0, t   e
  R L t



  R L t
W  , t   J  , t   e
(22)

 E1  t   R1 (W  , t   J  , t ) e

 R L s
 C11   W  , s   J  , s   e  
T
t


dL1  i  , t  
 i  , t 
 L1  i  , t   
di



d
W  , t   J  , t   e  R L t
dt
L0  t   i  0, t 
di

 R L t
dL1  i  , t  
di
Copyright © 2011 SciRes.
 L1  i  , t  

 2 L 

t
 C11   W  ,   J  0,  T   e ( R / L ) 2 L d
T
d
 L1  t 
W  , t   J  0, t  T   e ( R / L )t 2 L  .
dt

Then we put t  T  t and change the variables in the
integrals:




(23)
2 C 
t

 C01  W    J   T   e  R L  T 
T
 L0  t  T 


W  t  T   J  t   e  R L t T 

2 C 
 E1  t  T   R1 W  t  T   J  t   e
T

dL0  i  0, t  


 2 L 
 2 L   d
d
W  t   J  t  T   e  R L t T 
dt
  R L  t T 
 C11  W   T   J    e  R L  T 
 2 L   di
 L  (W  0, t   J  0, t ) e    2 L   ,
0
2 C 
 E0  t  T   R0 W  t   J  t  T   e   R L  t T 
 L0  i  0, t  


W  t   J  t  T   e  R L t T 
2 L .
  R L t


W  , t   J  0, t  T   e ( R / L )t
t
 W  0, t   J  0, t   e
L1  t   i  , t 


 2 L 
 2 L  ds 
 2 C 
 R L t
t

 R L
 C01   W  0, t  T   J  0,   e   2 L d 
T

d
 L0  t 
W  0, t  T   J  0,   e  R L 2 L
dt

Introduce denotations
dL0  i  0, t  

 E0  t   R0 W  0, t  T   J  0, t   e
 E1  t   R1 W  , t   J  0, t  T   e  ( R / L )t
2 C 
  R L t
2 C 
W  0, t  T   J  0, t   e R Lt
and
Replacing expressions (21) into the second boundary
condition we obtain the following equation

We assume that the unknown functions are W  , t 
 W  t  , J  0, t   J  t  . An integration along the charateristics yields W  , t  T   W  0, t  , J  , t   J  0, t  T 
and then Equations (22) and (23) become



1
 R L t
 2 L 
t

 R L t
 C   W  0,   J  0,   e   2 L d 
T



dL0  i  0, t  
  i  0, t  
 L0  i  0, t   
di


d
 R L t
W  0, t   J  0, t   e   2 L .


dt
1
0
  R L t
1
2 C 
  R L t
dL i  , t 
 2 L    di 
 L  W  , t   J  , t   e    2 L   .
 W  , t   J  , t   e
 2 L  ,
 2 C 
 2 L  d
d
 L1  t  T 
W  t  T   J  t   e  R L t T 
dt
 2 L .
To solve the above equations with respect to the derivatives dW (t ) dt and dJ (t ) dt we have to divide
the above equations into L0 (t  T ) and L1 (t  T ), respectively.
But
CS
V. G. ANGELOV
u k 
 di
d k dLk  i  dLk  i  .i  dLk  i 


 i
 Lk  i  
dt
dt
dt
d
i

 dt
 k  0,1
m
where Lk  i    an  i n . Then since
k
We also briefly recall:
i
t
dq d  C0  u  .u 

  i  x,  d  C0  u  .u  C0  u 
dt
dt
T
and
n0
m
m
t

uC0  u  0, t   C01   i  0,  d 
T

 k  i   i.Lk  i   i. an  i n   an  i n 1 .
k
n 0
we get
dLk  i 
i
di
dLk  i 
di
m
k
n 0
where ck ,  k , h  [2,3],
m
 i  nan  i n 1   an  i n
n 0
k
Introduce denotations
1 W , J  t   e
  R L  t T
2 L  ,

W  t  T   J  t    2 L  .
dL0   0 W , J  t  
L0 (t  T )   0 W , J  t 
 L0   0 W , J  t  
n0
n
 T
0
0
 T
0
0
imply
m
k 
Lk  t  T     n  1 an
n 0
  k W , J  t  
n
 Lk  0
 1 Lk  t  T   1 Lk  1 L  k  0,1
where 1 L  max 1 Lk : k  0,1 .
This can be done if the polynomials have suitable
properties (cf. 4. Numerical example).
Copyright © 2011 SciRes.
1
h k  u h
dCk  u 
du
1
, u   W0 , W0 .
 0 the inverse function exists for
ck u h  k
h
k  u
: W0 ,W0 
then
W  J e   2 L   i , (24)
W e  J   2 L   i
0
du
 k  h k  hu  u 
 c h  k W0 ck h  k W0 
,
 k

h  W
 h  k  W0
k
0 

We have to ensure a strict positive lower bound for
Lk  t  T  ,  k  0,1 . We can find an interval i  i0
such that the inequalities
   R L  T0
dCk  u 
Ck  u   uCk  u  
n 0
1 W , J  t   e
ck
h
di
m
0
 Ck  u   u

dL1  1 W , J  t  
   n  1 an1  1 W , J  t   .
   R L  T0
 k  0, 1
h k
.
h 1
Since
n
 L1  1 W , J  t  
k  u
u 
   n  1 an(0)   0 W , J  t   ,
L1  t  T   1 W , J  t 
du
Since
di
m
 0 W , J  t   e
dCk  u 
W  t   J  t  T  
Therefore
h
The derivatives of the functions Ck (u )  uCk (u ) are
n 0
 0 W , J  t   e
1  u / k 
ck h  k
u  W0    min  0 , 1   h  h  1   .
m
  R L  t T 
h

are constants and
n 0
  n  1 an(k )i n ,  k  0,1 .

ck
Ck  u  
 Lk  i 
k
301
ET AL.
 c h  k W0 ck h  k W0 
,
Ck1  I  :  k
   W0 , W0  ,
h  W
 h  k  W0

0 
k
or Ck1  I   W0 .
The explicit form of the inverse function for h  2 is:
ck  k u
k  u  I
 c k u  k I 2  I 2u
2
k
2
 ck2  k u 2  I 2 u   k I 2  0,
Ck1  I 

 I 2  I 4  4ck2  2k I 2
2ck2  k
 k  0,1 .
We need the derivative given below (see Equation (26))
CS
302
V. G. ANGELOV
and the following estimates:
I 
ET AL.
  W    J   T  e
t
 c  k W0
 min  k
,
h
 k  W0
  k  W0
ck  k W0
ck  k W0 

h  W

k
0 
h
h
  R L  T 
T
  W   T   J   e
t
and
  R L  T 
T
dCk1  I 
dI
I 2  4ck2  2k
1 

I

ck2  k 
I 2  4ck2  2k

1
I  I 2  4ck2  2k
 2
ck  k







  W   T   J   e
t
T

 W0  J 0 e T




2

2 2 
 4ck  k 


  R L  T 
T
  W   T   J   e
  R L  T 
T


 1  I 
2
dCk1  I   ck  k 

dI
 1 
 2  I 
 ck  k 
dt



dJ  t 
dt



dJ  t  T 
dt
2e 

R / L  t T 
L0
2e 
R / L  t T 
L0
dW  t  T 
dt

Copyright © 2011 SciRes.
L1
  R L  T 
  2 L     e
 2 L   d
   R L  T0
    R L 

1
For the I-V characteristics we assume that they are
polynomial functions
 2 L  d
Rk  i    rn  i n ,  k  0,1 .
m
k
n 1
Now we are able to formulate a periodic problem for
the obtained neutral system:
 c  k W0 
I 2  2ck2  k2 
, I   k
,0
2
2 2 

  k  W0
I  4ck  k 

,

ck  k W0 
I 2  2ck2  k2 
 , I   0,


 k  W0 
I 2  4ck2  k2 

(26)

R
R
R / L t T
W  t   J  t  T   2e   LE0  t  T   W  t   J  t  T   Z 0
L
L
  R / L  t T 


L
e
R0 
W t  J t  T  
 2 L   



  R / L  t T 
t

L 1  e
C0  
W    J   T   d   U W , J  t  .

T 2 L


L0
(27)
R
R
R L t T
W  t  T   J  t   2e   LE1  t  T  L1
L
L
0
  R L  t T 
 k  W0
 2 L  d ,.
 W  t  T   J  t   Z 
2e
ck  k W0
 R L T
W0 e  T  J 0 e    0  1 ck  k W0
.

   R L
 k  W0
2 L
of C01 . , C11 . should satisfy the inequalities
dW  t 
,
and
Therefore the arguments
t

 k  W0
 R L T
W0  J 0 e T e    0  1 ck  k W0

   R L
2 L
 k  W0
(25)
  W    J   T  e
 2 L   d
ck  k W0
we obtain
W
2
1  0  k  0,1 .
ck
k
t

or in view of the inequality

 c  k W0
1  ck  k W0
 2 
  k
  W
ck  k   k  W0
k
0



 2 L   d
L
2e
L1 
 R L  t T 
L1
L
R1
e
  R L  t T 
2 L
W  t  T   J  t  
(28)
 t e  R L  t T 

C11  
W   T   J    d   I W , J  s  , t  T , T  T0  ,

T 2 L

CS
V. G. ANGELOV
Bu W , J  t  :  0  t  , t   0, T  ,
W  t   0  t  , W  t   0  t  ,
J  t     t  , J  t     t  , t   0, T  .
0
Bi W , J  t  : 0  t  , t   0, T ，
0
The initial functions 0  t  , 0  t  can be obtained from
the initial conditions (7) by shifting along the characteristics of the initial functions u0  x  , i0  x  (cf. [9-11]).
t
Bu W , J  t  :   U W , J  s  ds
T
 t  T 1  T T0

   U W , J  s  ds
2 T
 T0
3. Main Results
Here we formulate conditions for the existence-uniqueness
of a periodic solution of neutral functional differential
system (27), (28) (for definition of neutral equation see
[15]). First we define a suitable operator generating by
the right-hand sides of the Equations (27), (28). We find
a T0  periodic solution of Equations (27), (28) on the
interval T , T  T0  coinciding with prescribed T0  periodic function on  0,T  and then one can continue it periodically because our system is autonomous one.
By L1,T0 T ,   we mean the space consisting of all
measurable essentially bounded T0  periodic functions
whose derivatives are also essentially bounded and T0 
periodic.
Introduce the sets:

M u  f  L1,T0 T ,   :
T T0
t   T , T  ,

M i  f  L1,T0 T ,   :
t   T , T  .

303
ET AL.

1
T0
T  T0 t
  U W , J  s  dsdt , t  T , T  T0 
T
T
t
Bi W , J  t  :   I W , J  s  ds
T
 t  T 1  T T0

   I W , J  s  ds
2 T
 T0
T T
1 0t
I W , J  s  dsdt , t  T , T  T0  .
T0 T T
Remark 1. It is easy to see that changing the integration
order one obtains

T T0 t
f  t  dt  0  f  t    0  t  ,

 U W , J  s  dsdt  T  T0 
T
T
T T0

U W , J  s  ds
T

T0
T T0

sU W , J  s  ds
T
and
T T0

f  t dt  0  f  t   0  t  ,
T
T T0 t
T T0
  I W , J  s  dsdt  T  T0   I W , J  s  ds
T
T
T

Lemma 1. If f (.)  M u (respectively f (.)  M i ) then
T T0

s I W , J  s  ds.
T
t
F (t )   f ( )d is T0  periodic function.
T
Assumption (E) Ek . L1,T0  T ,   ,
T T0

Ek  t dt  0,
T
Ek  t   W0 e
  t T 
, t  T , T  T0  , Ek  T   0  k  0,1 .
Lemma 3. The periodic problem (27), (28) has a solution W . , J .   M u  M i iff the operator B has a
fixed point W , J   M u  M i , that is, W  Bu W , J  ,
J  Bi W , J  .
Introduce the sets

f M
Assumption (IN) 0  t  ,0  t  are essentially bounded
T0  periodic functions where T  mT0 , m  2,3, 4,5,  ,
X u  f  M u : f  t   W0 e
0  0   0  0   0, 0 t  W0 , 0  t   J 0 . Here W0 , J 0
Xi
are prescribed positive constants.
Lemma 2. If the assumptions (IN) and (E) are fulfilled and W , J   M u  M i then R0   0 W , J  t   ,
t
R1  1 (W , J )(t )  ,  0  t      0 W , J    d
t
T
and 1  t     1 W , J    d are T0  periodic funcT
tions.
Define the operator B   Bu , Bi  on the set M u  M i
by the formulas:
Copyright © 2011 SciRes.
i
: f t   J0 e


, t  T , T  T   .
  t T 
  t T
, t  T , T  T0  ,
0
Here the constant   0 will be prescribed below.
The sets X u and X i turn out into metric spaces with
metrics


 W ,W   ess sup e   t T  W  t   W  t  : t  T , T  T0  ,


 (W , W )  ess sup e   (t T ) W (t )  W (t ) : t  [T , T  T0 ] ,


 ( J , J )  ess sup e   (t T ) J (t )  J (t ) : t  [T , T  T0 ] ,
CS
304
V. G. ANGELOV


 R L T
W0 e  T  J 0 e    0  1 ck  k W0
R

,  .
   R L
L
 k  W0
2 L
 ( J , J )  ess sup e  (t T ) J (t )  J (t ) : t  [T , T  T0 ] .
Define a metric on X u  X i in the following way:
  W , J  , W , J  

Theorem 1. Let the assumptions (LC), (E) and (IN)
be fulfilled. Then there exists a unique T0  periodic
solution W , J   X u  X i of the problem (27), (28).
Proof: Define the operator
B  ( Bu , Bi ) : X u  X i  X u  X i by the above formulas. In what follows we show
  

 max  W ,W  ,   J , J  ,  W ,W ,  J , J .
e
Assumption (LC):
e
W
   R L  T0
 J 0 e  T
0
W e
   R L  T0
0
 T
 J0
 i ,
0
2 L
 i ,
W (t )  W0 e  (t T )  J (t )  J 0 e  (t T )
0
2 L
 Bu  W0 e  (t T )  Bi  J 0 e  ( t T ) .
W0  J 0 e  T e   R L T0  1 ck  k W0

,
   R L
 k  W0
2 L
Bu (W , J )(t ) 
t
 U (W , J )( s )ds 
T

1
T0
ET AL.
1
2
Using inequality (24) and proceeding as in [11] we
obtain (for sufficiently large    R / L  ):
T  T0

U (W , J )( s )ds
T
T  T0 t
0
  0  1  T
 ( t T )  e
U
W
J
s
s
t
e
e J0
(
,
)(
)d
d



0
T


T


 T
(  0  2)e 0   0  2  R Z 0  W0  J 0 e
2 L W0 (e  T  1)  R L T0 (  0  2)e 0  2  0  2

e
 

L0    R L 

0
2 0
 L L0 

1

L0

n 1

rn(0) W0  J 0 e  T
m

n 2 L

n
 ( n 1)    R L  T0 2e 0  3 0  2 1  
 e 
    e  ( t T )W0 .
2

2  
0

     (n  1) R   nR  
n 1
For the second component of the operator we obtain
Bi (W , J )(t ) 
1
t
T T0
 I (W , J )( s)ds  2 
T
I (W , J )( s)ds 
T

1
T0
T T0 t
 

  I (W , J )( s)ds dt
T
T
 
 e  (t T ) W0 e  T e 0  0  1 0  W0 e  T  1 


  R L    Z 0 L1 

 2e
0

   0  2  e 
 30  2 /  2 0 L1   
m
0
 0  2

 2  0    4 e  R L T
0

rn(1) W0 e  T  J 0
n 1
2 L 
n 1
n

n
 e
   R L  T0

1


L W0 e T  1
L1   R L    

n  ( n 1)  R L  T0
1 
e
 J 0 e  ( t T ) .
    (n  1) R   nR   

Therefore the operator B maps the set X u  X i into itself. In what follows we show that B is contractive operator.
We need the following preliminary inequalities
 0 (W , J )(t )   0 W , J  (t ) 
1 (W , J )(t )  1 W , J  (t ) 
Copyright © 2011 SciRes.
e
   R L   ( t T )
2 L
e
   R L   ( t T )
2 L
  W ,W     J , J  e  ,
 T
  W ,W  e
 T
 
  J , J ,
CS
V. G. ANGELOV
   0 W , J  (t ) 
 0 (W , J )(t )
n
 n  e

   R L  T0
W
0
1 (W , J )(t )n  1 W , J  (t ) 
 n  e

   R L  T0
W e
 T
0
n
  2 L 
 J 0 e  T
 J0
305
ET AL.
n 1
 e
   R L   ( t T )
  W ,W     J, J  e   2 L  .
 T
n
  2 L 
n 1
 e
   R L   ( t T )
  W ,W  e
 T
    2 L  .
  J , J
Then for the first operator component we obtain:
Bu W , J  (t )  Bu W , J  (t )
T  T0
t
 t  T 1  T T0
  U W , J  ( s )  U W , J  ( s ) ds  
   U W , J  ( s )ds   U W , J  ( s )ds
2 T
T
T
 T0

e
1
T0
T  T0 t
t

 U W , J  (s)ds   U W , J  (s)ds dt
T
T
T

0
 T
( 0  2)e 0  0  2 1  e  T
  W , J , W , J ) e  0  1 e






2 0
0

2

 ( t T )

1

   R L    Z 0 L0  
L0


r


m
n 1
(0)
n
ne
( n 1)     R / L  T0
(W0  J 0 e  T ) n 1
2 L 
n 1
 2 1  W0  0 
 c L     R L 
0
0





 e  (t T ) K u  W , J  , W , J  .
Further on we have
Bi (W , J )(t )  Bi (W , J )(t )
T T0
t
 t  T 1  T T0
  I (W , J )( s )  I (W , J )( s ) ds  
   I (W , J )( s )ds   I (W , J )( s )ds
2 T
T
T
 T0

e
1
T0
T T0 t
t

 I (W , J )(s)ds   I (W , J )(s)ds dt
T
T
 ( t T )
T
 0
e  0  1 e T ( 0  2)e 0  0  2 (e T  1)

 ((W , J ), ( J , W )) 
2 0
0

2



( n 1)     R / L  T0
(W0 e  T  J 0 ) n 1
2 m (1) ne
2


   R L    Z 0 L0    rn

1  W0 1   
n 1
L

c
L
R
L






n
1
1
1 1


2 L







 e  (t T ) K i  W , J  ,  J ,W  .
Finally we have to obtain an estimate for t  [T , T  T0 ] . It is easy to prove the following inequalities:
Copyright © 2011 SciRes.
CS
306
V. G. ANGELOV
ET AL.
Bu (W , J )(t )  Bu (W , J )(t )
 U (W , J )(t )  U (W , J )(t ) 
e
 ( t T )
1
T0
T  T0

U (W , J )( s )ds 
T
T  T0

U (W , J )( s )ds
T


 T

R Z
e 0   0  1 1  e
1
 T

  0 
 ((W , J ), (W , J )) e 
L
L
L



0
0
0




m

n 1
(0)
n
r
ne
( n 1)     R L  T0
W
0
2 L 
 J 0 e  T

n 1
n 1


L0 c0     R L    
2 1  W0  0

 e  (t T ) K u  W , J  ,  J , W  .
For the derivative of the second component of B we obtain
B (W , J )(t )  B (W , J )(t )
i
i
 I (W , J )(t )  I (W , J )(t ) 
1
T0
T  T0

I (W , J )( s)ds 
T
T  T0

I (W , J )( s )ds
T

e 0  0  1 (e T  1)
 e  (t T )  ((W , J ), (W , J ))  e T 
0


R Z
e
1
   0   n rn(1)
L
L
L

1
1 n 1
m

R

( n 1)    T0
L

(W0 e T  J 0 ) n 1
(2 L )
n 1

 e  (t T ) K i  W , J  ,  J , W  .
The above inequalities imply
 ( B(W , J ), B(W , J ))  K  ((W , J ), (W , J ))



c1 L1     R L    
2 1  W0 1
We choose   2 1012. Then T0  0  2 and
T  6 109  2 1012 T0  12000 T0 . Consequently
where K  max  K u , K i , K u , K i  should be chosen
T  2 1012  6 10
smaller than 1.
Therefore B is contractive operator and has a unique
fixed point in M (cf. [18]). It is a unique periodic solution
of system (26), (27).
Theorem 1 is thus proved.
R L  0.0029 (0.45 106 )  0.6 104 , e
 R L  T0  0  e( R / L )T
4. Numerical Example
   R L   1012 , e
For a transmission line with length   1m,
L  0.45  H m, C  80 pF m , cross-section area
S  6 mm 2 , specific resistance for the cuprum
c  0.0175 , that is, R   c   S  3.103  ,
v 1
LC  1 (6 109 )  1.66 108 ,
9
Z 0  L C  75  . Then T   LC  6 10 sec .
Let us check the propagation of millimeter waves
0  1 6  103 m . We have
f 0  1 0 LC  1012 Hz  T0  1 f 0  1012.


Copyright © 2011 SciRes.
9
 12 103  e T  0 , RT L  0,
 RT L
 e0  1
   R L   2 1012  104  1012 (2  108 )  2 1012 ,
0
 1  6.6 106 ,
   R L  T0
 e 0 e
  R L T0
 e2 .
We choose resistive elements with the following V-I
characteristics R0 (u )  R1 (u )  0.028u  0.125u 3 i.e.
r1  0.028, r2  0, r3  0.125 , and inductive element with
L0 (i )  L1 (i )  3 i  1 12  i 3 . Then
L (i )  i  dL (i ) di   L (i )  6 i  1 3 i 3 .
0
0
0
If we choose i0  1 one obtains
6i  1 3 i 3  6  1 3  17 / 3
and consequently 1 L  3 17 .
Let us take
C0 (u )  C1 (u )  c
1  u    c 
 u ,
where h  2, c  50 pF  5 1011 F and   0.4 V 
CS
V. G. ANGELOV
W0  0.4 . Then the inequalities of Theorem 1 become
e
W
    R L  T0
0
 e 2W0
e
 J0e
 T
0
 e J0
2

2 L
2
  R L T0
W0 e 2  1
 2 L   1
2 L   1  W  2 L e
W e  J   2 L   1
2 L   1  J  2 L e


cW0

  W0
J 0 e2  1
0
2 L
2
0
 0.18 10

  W0 
2c  L
e2  1

 W0  min 13.252  0.4; 0.18 103  0.18 103 ,
 0.18 103 ,
 T
0
307
ET AL.
3
 J0 
and



cW0
  W0
2  c L   W0 
(e 2  1)   W0
 6 106 ;
1  2  R Z 0  (4e2  2) L
1
e2  1 0,125W02 e6  2e 4  1  




e
0,
028

   1,

   L L 
L
L
2
L
24
 
1  R Z 0


  L L

(e2  3) L W0 e 2  2 
0,125 J 03 6
 2
2
e
W
0.028
J
(
e
1)
(e  1)    J 0 ,





0
0

L
2L 
12 L

 
3e4 W02
e2  1  R Z 0 1 
K u 
0.028
0.125




2  L L L 
4L
3e 4 J 02
e2  1  R Z0
1
K i 

  0.028  0.125

2  L L
L
4L

  2 1  (W0 / ) / (  Lc)   1,


  2 1  (W0 /  ) / (  Lc)   1.

Taking into account R L  Z 0 L  104 (0.6  13.24 10 4 )  0.6 104 , W0  0.18 103 and J 0  3 106 we see that
1/ (2 1012 )  4.44 104  5.58 0.45 106  13 / 17  0.118  3.4    1,




106 8  0.83 0.54 106  (3.375 10 12 )404 / 5.4   6,
K u , K i  K i  K u  (2,1/ 1012 )(0.6 104  0.072  0.0036)13 107  1.
Most often the initial approximation is chosen to be
simple functions, namely:
W sin 0 t , t  [0, T ]
W (0) (t )   0
;
 0, t  [T , T  T0 ]
 J sin 0 t , t  [0, T ]
J (0) (t )   0
 0, t  [T , T  T0 ]
 ((W
,J

 1.13 107
), (W
,J
( n)
7
t
(1)

t
 2 L  e
W (t )  J (t  T ) 
Copyright © 2011 SciRes.



3
R L  ( s T )

R0  0 (W (0) , J (0) )( s )
  L (s  T )  ds
0
 2 L 
R L  ( s T )
R0 (i0 )
 L (s  T )  ds
0
T
t
L0 (t  T )


, J (1) , W (0) , J (0) ,
We notice that L0 (i )  L1 (i )  6i  1 3 i 3 imply
 ( R / L )( t T )

T
(n  1, 2 )
6 e

 6 e ( R / L )(t T ) W (t  T )  J (t )  2 L
2 L  e
))
 1 1.13 10    W
n
3
and therefore (recall  0  i0  1, 1  i0  1 ):
J ( n 1) (t )  Bi (W ( n ) , J ( n ) )
(n  0,1, 2,) and
(n)
 ,
L1 (t  T )

Then we have W ( n 1) (t )  Bu (W ( n ) , J ( n ) ),
( n 1)

 1 3 e  ( R / L )(t T ) W (t  T )  J (t )  2 L
( 0  2π T0 ) and E0 (t )  E1 (t )  E0 sin 0 t .
( n 1)

 1 3 e ( R / L )(t T ) W (t )  J (t  T )  2 L
 2 L  e
R L  ( s T )
 0, 028  0,125 
( L )ds
T

 0.31L L e
R L T0
  LR   0.074 10
1
12
 0.
Analogously
CS
308
V. G. ANGELOV
t
2 L  e
R L  ( s T )

R1 1 (W (0) , J (0) )( s )

ET AL.




R LT
L1 ( s  T )ds  6 L 17  0.028  0.125  e  0  1 L R  0.
T
We find the first approximation
t
 t  T 1  T T0
   U (W (0) , J (0) )( s )ds
W (1) (t )  Bu (W (0) , J (0) )(t )   U (W (0) , J (0) )( s )ds  
T
2 T
T
 0

1
T0
T T0 t
  U (W
T
(0)
, J (0) )( s )dsdt
T
t
t
T
T
  J 0 sin 0 t   R L   J 0 sin 0 s ds  2 E0 L  e
R L  ( s T )


sin 0 s L0  s  T  ds
t
t  R L  ( s T )

C01  s   R L  ( T )
e
 Z 0 J 0  sin 0 s L0  s  T  ds  2 L 
J 0 sin 0 d ds
e

T
T L0 ( s  T ) 2 L  T



T T0
T T0
 t T 1 
R L ( s T )
sin 0 s L0  s  T  ds  Z 0 J 0  sin 0 s L0  s  T  ds

   2 E0 L  e 
2  
T
T
 T0

2 L
T T0

T

1
T0
T  T0

T



R L ( s T )
 J s  R L ( T )
 
e 
C01  0  e  
sin 0 d ds 

L0 ( s  T )
2 L T
 
t
t

 R L  ( s T )
sin 0 sL0 ( s  T )ds
  J 0 sin 0 t   R L   J 0 sin 0 sds  2 E0 L  e
T
T

t
t  R L  ( s T )
 J s  R L ( T )
 
e
C01  0  e  
sin 0 d ds  dt.
 Z 0 J 0  sin 0 s L0  s  T  ds 2 L 

T
T L0 ( s  T )
2 L T
 

t
Since
 R L   J 0 sin 0 s ds

 J 0 109 and
T
R L ( s T )
s

2 L L  R L T0
e 
  R L  ( T )
1  J 0
 1  W0 3.6 108  0
sin 0 d ds  W0
C
e
 L (s  T ) 0  2 L  e
L
R
T 0
T



t
2 L

we can disregard this terms and obtain
t
W (1) (t )   J 0 sin 0 t  2 E0 L 
e
R L  ( s T )
t
sin 0 s
ds  Z 0 J 0  sin 0 s L0 ( s  T ) ds
L0 ( s  T )
T
T


T T0
T T0

 t T 1 

   2 E0 L  e R L ( s T ) sin 0 s L0 ( s  T ) ds  Z 0 J 0  sin 0 s L0 ( s  T ) ds 


2 
T
T
 T0



1
T0
T T0

T



t
t


 R L  ( s T )
sin 0 s L0 ( s  T ) ds  Z 0 J 0  sin 0 s L0 ( s  T ) ds  dt ,
 2 E0 L  e
T
T






and then
W (1) (t )  W (0) (t )


 J 0  2 E0 L L RL e
R L T0



t
 1   Z 0 J 0 L   e  ( s T ) ds  E0 L L RL e
R L T0

1
T
  Z0 J 0 2L 
T T0

e  ( s  T ) ds 
T


 5 L3 2 LR e
R L T0

1
T0
T T0

T
 2E L
t

L RL e R L T0  1   Z 0 J 0 L   e  ( s T ) ds  dt  J 0

T


0


  2L  
 1 E0  5Z 0 J 0 e 0  1
 e 0 J 0  e 2 J 0 .
Copyright © 2011 SciRes.
CS
V. G. ANGELOV
309
ET AL.
For the derivative we have
1
W (1) (t )  U (W (0) , J (0) )(t ) 
T0
  0 J 0 cos 0 t 
T T0

U (W (0) , J (0) )( s )ds
T
E0 L
2e 
R
J 0 sin 0 t 
L
L0 (t  T )
R L ( t T )
2e
Z
 0 J 0 sin 0 t 
L0
 R L  ( t T )

L R  0 (W (0) , J (0) )(t )
L (t  T )

0

T T0
R L ( t T )
 1
L 1  t
2e 
R L ( s T )
C0    0 (W (0) , J (0) )( )d    2 E L  e 


L0 (t  T )
T
T
 T0 

T T0
1
 Z 0  J 0 sin 0
T0 
T

 L (s  T )  ds 2
0
T T0
L

e
R L  ( s T )
L0 ( s T )

ds 



R0  0 (W (0) , J (0) )( s )
T

  L (s  T )  ds 
0

T T0  R L  ( s T )
s
 
e
1
L
2
C01    0 (W (0) , J (0) )( )d  ds  .




T0 
T L0 ( s  T )
T
 
Then we obtain
W (1) (t )  W (0) (t )


Z
R Z


  0   0  0 (e 0  1) 0  J 0  2 L L e RT0
L
L
L





 2 L L  0.028  0.125 e RT0
Since
e
RT0 L

1
 RT0 L   1,
L
 (e RT0
L
L

 e RT0
 
L

1
 RT0 L    E0  W0 

 1)  RT0 L   2 L L e RT0
L

1
 RT0 L W0 .
E0  0.5 V we have

225(e2  1) 
3 0.45 106
W (1) (t )  W (0) (t )   2π 1012  0, 6.104 
 2  6W0  0, 612   3.768 107.
 J0 
34
17


(1)  (0)
6

It follows  W ,W
 18.84 10 . Consequently


1.13 10 
)) 
7
 ((W
( n 1)
,J
( n 1)
), (W
( n)
,J
(n)
n
1  1.13 107
3.8 107 (n  0,1, )
W ( n 1) (t )  W ( n ) (t )  7.4(1.13 107 ) n 3.8 107  (1.13 107 ) n  (2.82) 108.
or
In the same way we can obtain an estimate for the second component of the operator B:
J (1) (t )  Bi (W (0) , J (0) )(t ) 
 W0 sin 0 t 
T T
 t  T 1  T T0
1 0t
(0)
(0)
(0)
(0)



I
(
W
,
J
)(
s
)d
s
I
(
W
,
J
)(
s
)d
s
I (W (0) , J (0) )( s )dsdt



2  T
T0 T T
T
 T0
t
t
RW0 t
e R L ( s T )

sin
d
2
ds

s
s
E
L
0
0

L T
T L1 ( s  T )
T T0  R L  ( s T )
T T0

 t T 1 
W sin 0 s
W0 sin 0 s
e
   2 E0 L 
ds  
ds  Z 0  0
ds 



2  
L1 ( s  T )
T L1 ( s  T )
T L1 ( s  T )
T
 T0

t
 Z0 

1
T0
T T0

T
t  R L  ( s T )
t


W0 sin 0 s
e
E
L
2
 0
 L (s  T ) ds  Z 0  L (s  T ) ds  dt.
T 1
T
1


Copyright © 2011 SciRes.
CS
310
V. G. ANGELOV
5. Acknowledgements
The research described in this paper was carried out
within the frame work of Contract No. DUNK-01/03
-12.2009 (UNIK).
6. References
[1]
M. Shimura, “Analysis of Some Nonlinear Phenomena in
a Transmission Line,” IEEE Transactions on Circuit
Theory, Vol. 14, No. 1, 1967, pp. 60-68.
doi:10.1109/TCT.1967.1082648
[2]
L. O. Chua and P.-M. Lin, “Machine Analysis of Electronic Circuits,” in Russian, Energy, Moscow, 1980.
[3]
D. M. Pozar, “Microwave Engineering,” 2nd Edition,
New York, Wiley, 1998.
[4]
P. Chirlian, “Analysis and Design of Integrated Electronic Circuits,” Harper & Row, Publishers, Inc., New
York, 1987.
[5]
C. R. Paul, “Analysis of Multiconductor Transmission
Lines,” Wiley-Interscience Publications, John Wiley &
Sons, Hoboken, 1994.
[6]
L. O. Chua, C. A. Desoer and E. S. Kuh, “Linear and Nonlinear Circuits,” McGraw-Hill Book Company, New
York, 1987.
[7]
A. H. Zemanian, “Hyperreal Waves on Transfinite, Terminated Distortionless and Lossless Transmission Lines,”
International Journal of Circuit Theory and Applications,
Vol. 33, No. 3, 2005, pp.183-193.
doi:10.1002/cta.314
[8]
S. Rosenstark, “Transmission Lines in Computer Engineering,” McGraw-Hill, Inc., New York, 1994.
[9]
V. G. Angelov, “Transmission Lines Terminated by Series Nonlinear RLC-Loads,” Annual of the University of
Architecture, Civil Engineering and Geodesy, Vol. 41,
Sofia, 2000-2001, pp. 9-29.
[10] V. G. Angelov, “Lossy Transmission Lines Terminated
by Nonlinear R-Loads with Exponential V-I Chara- cteristics,” Journal Nonlinear Analysis, Vol. 8, No. 2, 2007,
pp. 579-589.
Copyright © 2011 SciRes.
ET AL.
[11] V. G. Angelov, “Periodic Regimes for Distortionless Lossy
Transmission Lines Terminated by Parallel Connected
RCL-Loads,” In: Transmission Lines: Theory, Types and
Applications. Ser. Electrical Engineering Developments,
Nova Science Publishers, New York, 2011.
[12] S. Grivet-Talocia, H. M. Huang, A. E. Rudehli, F. Canavero and I. M. Elfadel, “Transient Analysis of Lossy
Transmission Lines: An Efficient Approach Based on the
Method of Characteristics,” IEEE Transactions on Advanced Packaging, Vol. 27, No. 1, 2004, pp. 45-56.
doi:10.1109/TADVP.2004.825467
[13] V. Rasvan, “Functional Differential Equations and OneDimensional Distortionless Propagation,” Tatra Mountains Mathematical Publications, Vol. 43, 2009, pp. 215228.
[14] K. L. Cooke and D. W. Krumme, “Differential-Difference Equations and Nonlinear Initial Boundary Value
Problems for Linear Hyperbolic Partial Differential Equations,” Journal of Mathematical Analysis and Applications, Vol. 24, No. 2, 1968, pp. 372-387.
doi:10.1016/0022-247X(68)90038-3
[15] A. D. Myshkis, “Linear Differential Equations with Retarded Arguments,” in Russian, Nauka, Moscow, 1972.
[16] M. A. Krasnoselskii, “The Shift Operator along the Trajectories,” in Russian, Moscow, 1970.
[17] Y. G. Borisovich, “Nonlinear Fredholm Mappings and
Periodic Solutions of Functional Differential Equations,”
in Russian, Proceedings of Mathematical Faculty, Vol.
10, 1973, pp. 12-25.
[18] V. G. Angelov, “Fixed Points in Uniform Spaces and
Applications,” Cluj University Press, Cluj-Napoca, 2009.
[19] T. Dhaene and D. D. Zutter, “Selection of Lumped Models for Coupled Lossy Transmission Lines,” IEEE
Transaction on Computer-Aided Design, Vol. 11, No. 7,
1992, pp. 805-815. doi:10.1109/43.144845
[20] S. Y. Lee, A. Konrad and R. Saldanha, “Lossy Transmission Line Transient Analysis by the Finite Element
Method,” IEEE Transactions on Magnetics, Vol. 29, No.
2, 1993, pp. 1730-1732. doi:10.1109/20.250739
[21] Taflove and S. C. Hagness, “Computational Electromagnetics: The Finite-Difference Time-Domain Method,” 2nd
Edition, Artech House, Boston, 2000.
CS